Problem: Leaving the outpost, Alejandro went on a three-day journey into the desert. His displacement (distance and direction) from start to finish on day one was ${\vec{d_1}} = 3\hat{i} + 7\hat{j}$. His displacement from start to finish on day two was ${\vec{d_2}} = -1\hat{i} + 8\hat{j}$. His displacement on day three, ${\vec{d_3}}$, took him back to the outpost. (Distances above are given in kilometers, $\text{km}$.) What distance did Alejandro travel on day three?
In order for ${\vec{d_3}}$ to get Alejandro back to the outpost, his horizontal and vertical displacements over the course of the three days must both sum to zero. By this logic, if ${\vec{d_1}} + {\vec{d_2}}$ is ${( 3 \hat i + 7 \hat j )} + {( -1 \hat i + 8\hat j )} = 2 \hat i + 15\hat j$, then ${\vec{d_3}}$ must be $-2 \hat i + (-15)\hat j$. Graphically, we can picture this situation as follows: We can find the magnitude of ${\vec{d_3}} $ using the Pythagorean theorem, which will tell us the distance Alejandro walked back to the outpost on day three. $\begin{aligned} \|{\vec{d_3}} \|^2 &= -2^2 + (-15)^2\\\\ \|{\vec{d_3}} \| &= \sqrt{4 + 225}\\\\ \|{\vec{d_3}} \| &= \sqrt{229}\\\\ \|{\vec{d_3}} \| &\approx 15.1 \text{ km} \end{aligned}$ Finding the direction of ${\vec{d_3}}$ will tell us in what direction Alejandro walked on day three. ${\vec{d_3}}$ is pointing in the third quadrant with an $x$ -component of $-2$ and a $y$ -component of $-15$. We can find the direction of any vector in the third quadrant using the arctangent function and adding $180^\circ$. $\begin{aligned} \tan \theta &= \dfrac{y}{x}\\ \\ \tan \theta &= \dfrac{-15}{-2}\\\\ \theta &= \arctan{\left( \dfrac{15}{2} \right)} \\\\ \theta&\approx 82^\circ \end{aligned}$ Adding $180^\circ$ we get $262^\circ$. Alejandro traveled $15.1$ kilometers on day three. Alejandro traveled in a direction of $262^\circ$ on day three.